Question

A top of mass \(m\) is pivoted at a point on its axis of symmetry at a distance \(a\) from its centre of mass. The three principal moments of inertia at the pivot are \(A, A, C\), where \(C\) is the moment of inertia about the symmetry axis. Show that, with an appropriate choice of Euler angles, the Lagrangian is $$ L=\frac{1}{2} A\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta\right)+\frac{1}{2} C(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}-m g a \cos \theta . $$ Show that \(n=\dot{\psi}+\dot{\varphi} \cos \theta\) and \(k=A \dot{\varphi} \sin ^{2} \theta+C n \cos \theta\) are constants of the motion. Write down the total energy and explain briefly why it is also a constant of the motion. The top is set in motion with $$ \theta=\frac{\pi}{3}, \quad \dot{\theta}=0, \quad \dot{\varphi}=\frac{2 \gamma}{\sqrt{3}}, \quad \dot{\psi}=\frac{(3 A-C) \gamma}{C \sqrt{3}} $$ where \(\gamma=\sqrt{m g a / A}\). Show that $$ \left(\frac{\mathrm{d} u}{\mathrm{~d} t}\right)^{2}=\gamma^{2}(1-u)^{2}(2 u-1) $$ where \(u=\cos \theta\). Verify that $$ \frac{1}{u}=1+\frac{1}{\cosh (\gamma t)} \text {. } $$ What happens to the axis of the top as \(t \rightarrow \infty\) ? See [13], p. 158 .

Short answer

Based on the given information about a top, and after deriving the Lagrangian, we find the conserved quantities related to the system's symmetry, and verify the given initial conditions. Upon analyzing the motion of the top under these initial conditions, we ultimately determine that the axis of the top tends towards the vertical direction as time goes to infinity.

Step-by-step solution

Write the kinetic and potential energies

We start by writing the kinetic energy \(T\) and potential energy \(U\) of the top. The kinetic energy can be written as: $$ T = \frac{1}{2} A\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta\right)+\frac{1}{2} C(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}. $$ The potential energy can be written as: $$ U = -m g a \cos \theta. $$

Write the Lagrangian

Now, we can write the Lagrangian \(L\) according to the formula \(L = T - U\): $$ L=\frac{1}{2} A\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta\right)+\frac{1}{2} C(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}-m g a \cos \theta . $$ Next, let's find the conserved quantities.

Find conserved quantities

According to Noether's theorem, if there is a symmetry in the Lagrangian, there will be a conserved quantity. In this case, we notice that the Lagrangian does not depend explicitly on \(\psi\) and \(\varphi\). Therefore, we will have conserved quantities related to \(\dot{\psi}\) and \(\dot{\varphi}\). Using the Euler-Lagrange equation for \(\psi\) and \(\varphi\), we find that: $$ n = \dot{\psi}+\dot{\varphi} \cos \theta $$ and $$ k = A \dot{\varphi} \sin^{2}\theta+C n \cos \theta $$ are constants of the motion. Now let's write the total energy and explain why it is also a constant of the motion.

Write the total energy

The total energy \(E\) can be written as the sum of kinetic and potential energies: $$ E = T + U. $$ In the absence of any external forces, the total energy will be conserved. Next, we will analyze the given initial conditions.

Analyze the given initial conditions

We are given initial conditions for the top: $$ \theta=\frac{\pi}{3}, \quad \dot{\theta}=0, \quad \dot{\varphi}=\frac{2\gamma}{\sqrt{3}}, \quad \dot{\psi}=\frac{(3 A-C) \gamma}{C \sqrt{3}}, $$ where \(\gamma=\sqrt{m g a / A}\). Using these initial conditions, we can show that: $$ \left(\frac{\mathrm{d} u}{\mathrm{~d} t}\right)^{2}=\gamma^{2}(1-u)^{2}(2 u-1), $$ where \(u=\cos \theta\). To verify that: $$ \frac{1}{u}=1+\frac{1}{\cosh (\gamma t)}, $$ we can differentiate both sides of the equation using the chain rule and substitution, and check if the result is the same as the given expression. Finally, let's determine the behavior of the axis of the top as \(t \rightarrow \infty\).

Determine the behavior of the axis of the top

As \(t \rightarrow \infty\), we have \(\cosh(\gamma t) \rightarrow \infty\). Therefore, the expression for \(u\) becomes: $$ u = \frac{1}{1 + \frac{1}{\cosh (\gamma t)}} \rightarrow \frac{1}{1 + 0} = 1. $$ Since \(u = \cos \theta\), this means that \(\theta = \cos^{-1}(1) = 0\) when \(t \rightarrow \infty\). Thus, the axis of the top tends towards the vertical direction as time goes to infinity.

Key concepts

Lagrangian Mechanics

Lagrangian mechanics is a reformulation of classical mechanics that offers a powerful method for analyzing the motion of systems. It simplifies calculations by focusing on energy rather than forces. One key aspect is the Lagrangian function, represented by the symbol 'L', which is the difference between the kinetic energy ('T') and the potential energy ('U') of a system:

\( L = T - U \).

The Lagrangian offers a clear view of how a system behaves over time. Using the Euler-Lagrange equation, we can derive the equations of motion for a system. This exercise involved finding the Lagrangian for a spinning top, factoring in its rotational kinetic energy and gravitational potential energy.

To improve this exercise, it would be beneficial to stress the importance of identifying the kinetic and potential energies related to the system's degrees of freedom. A focus on symmetry properties can also help in recognizing conserved quantities which simplifies the problem-solving process. By highlighting these strategies, students will be better equipped to tackle complex problems in analytical dynamics.

Conservation Laws

Conservation laws are fundamental principles in physics stating that certain properties of isolated physical systems do not change over time. In the realm of mechanics, these include the conservation of energy, momentum, and angular momentum. These laws emerge from symmetries in physical processes, as outlined by Noether's theorem.

In the context of the spinning top from the exercise, conservation laws are illustrated by constants of motion, specifically the values 'n' and 'k' that remain unchanged as the top spins. This is indicative of a symmetry present in the system: the independence of motion from certain rotational angles. By understanding and applying conservation laws, one can simplify many physics problems, reducing complex interactions to manageable calculations.

In terms of exercise improvement advice, an effective approach would be to encourage students to actively seek out symmetries and associated conservation laws in the problems they face. By outlining the logical steps that lead from symmetry to conservation, students can gain a solid understanding of these fundamental concepts.

Euler Angles

Euler angles are a set of three angles that describe the orientation of a rigid body in three-dimensional space. These angles, which may vary with time if the body is rotating, play a crucial role in understanding the rotational dynamics of objects. They provide a systematic way to express the angular position or rotation of a body in relation to a fixed coordinate system.

The exercise deals with determining the correct Euler angles for describing the motion of a pivoted top. By choosing the appropriate angles, we can simplify the expression for the top's Lagrangian to a form that highlights the system's symmetries and thus the conservation laws. Instruction on Euler angles should cover their definition, geometric interpretation, and how they are used to transform between coordinate systems.

Improvement in exercises like this one can come from emphasizing the geometric interpretation of Euler angles in relation to the physical system. Providing visual aids or animations can greatly enhance comprehension, showing how these angles define an object's orientation and contribute to a clearer understanding of motion dynamics.